# How do you solve -3x + 5y + z = 10, 2x + 3y - z = 7, and -4x + 2y +3z = -1 using matrices?

Dec 29, 2016

x=-1, y=2, z= -3

#### Explanation:

In matrix form, the system can be written as follows and the solved using indicated row operations:

$\left[\begin{matrix}- 3 & 5 & 1 \\ 2 & 3 & - 1 \\ - 4 & 2 & 3\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}10 \\ 7 \\ - 1\end{matrix}\right]$

${R}_{1} + {R}_{2} \to {R}_{1}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 2 & 3 & - 1 \\ - 4 & 2 & 3\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 7 \\ - 1\end{matrix}\right]$

${R}_{3} + 2 {R}_{3} \to {R}_{3}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 2 & 3 & - 1 \\ 0 & 8 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 7 \\ 13\end{matrix}\right]$

$2 {R}_{1} + {R}_{2} \to {R}_{2}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 0 & 19 & - 1 \\ 0 & 8 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 41 \\ 13\end{matrix}\right]$

${R}_{2} - 2 {R}_{3} \to {R}_{2}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 0 & 3 & - 3 \\ 0 & 8 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 15 \\ 13\end{matrix}\right]$

${R}_{2} / 3 \to {R}_{2}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 0 & 1 & - 1 \\ 0 & 8 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 5 \\ 13\end{matrix}\right]$

${R}_{3} - 8 {R}_{2} \to {R}_{3}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 9\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 5 \\ - 27\end{matrix}\right]$

${R}_{3} / 9 \to {R}_{3}$

$\left[\begin{matrix}- 1 & 8 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}17 \\ 5 \\ - 3\end{matrix}\right]$

$- 1 {R}_{1} \to {R}_{1}$

$\left[\begin{matrix}1 & - 8 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}- 17 \\ 5 \\ - 3\end{matrix}\right]$

${R}_{3} + {R}_{2} \to {R}_{2}$

$\left[\begin{matrix}1 & - 8 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}- 17 \\ 2 \\ - 3\end{matrix}\right]$

$8 {R}_{2} + {R}_{1} \to {R}_{1}$

$\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right] \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}- 1 \\ 2 \\ - 3\end{matrix}\right]$